已知函数f(x)=(1-tanx)[1+√2sin(2x+π⼀4)],求函数定义域值域，最小正周期，单调递增区间

已知函数f(x)=(1-tanx)[1+(√2)sin(2x+π/4)],求函数定义域值域，最小正周期，单调递增区间
解：定义域：x≠kπ+π/2
f(x)=[(cosx-sinx)/cosx](1+sin2x+cos2x)=[(cosx-sinx)+(cosx-sinx)(sin2x+cos2x)]/cosx
=[(cosx-sinx)+sin2xcosx-sin2xsinx+cos2xcosx-cos2xsinx)]/cosx
=[(cosx-sinx)+(sin2xcosx-cos2xsinx)+(cos2xcosx-sin2xsinx)]/cosx
=[(cosx-sinx)+sin(2x-x)+cos(2x+x)]/cosx
=(cosx-sinx+sinx+cos3x)/cosx=(cosx+cos3x)/cosx=2cos2xcosx/cosx=2cos2x
故最小正周期T=π；
单调递增区间：由-π+2kπ≦2x≦2kπ，得单增区间为-π/2+kπ≦x≦kπ，k∈Z

).cosx≠0.
x≠kπ+π/2.k∈Z.
∴f(x)的定义域为{x│x≠kπ+π/2.k∈Z}.

2)α是第四象限的角，切tanα=-4/3.
sinα=-4/5, cosα=3/5.
sin2α=2sinαcosα=-24/25.
cos2α=2cos^2-1=-7/25.
√2sin(2x-π/4）=√2(sin2xcosπ/4-cos2xsinπ/4)
=sin2x-cos2x
=-17/25.
∴f(α)=1-(-17/25)/(3/5)
=32/15.

由f(x)=(1-tanx)[1+√2sin(2x+π/4)]化简得：

函数的值域是[0,4],最小正周期为π,单调增区间为[Kπ-π/4,Kπ+π/4]